If alpha , beta neq 0 and f(n) = alpha^n + be | vedclass

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(A) Given $f(n) = \alpha^n + \beta^n$. The determinant is $\Delta = \begin{vmatrix} 1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \ 1+\alpha+\beta & 1+

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$1$

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(A) Given $f(n) = \alpha^n + \beta^n$. The determinant is $\Delta = \begin{vmatrix} 1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \ 1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \ 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4 \end{vmatrix}$.This can be written as the product of two determinants: $\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & \alpha & \beta \ 1 & \alpha^2 & \beta^2 \end{vmatrix} 	imes \begin{vmatrix} 1 & 1 & 1 \ 1 & \alpha & \beta \ 1 & \alpha^2 & \beta^2 \end{vmatrix}$.Each determinant is a Vandermonde determinant,which equals $(1-\alpha)(1-\beta)(\alpha-\beta)$.Therefore,$\Delta = [(1-\alpha)(1-\beta)(\alpha-\beta)]^2 = (1-\alpha)^2 (1-\beta)^2 (\alpha-\beta)^2$.Comparing this with $K(1-\alpha)^2 (1-\beta)^2 (\alpha-\beta)^2$,we get $K = 1$.

If $\alpha , \beta \neq 0$ and $f(n) = \alpha^n + \beta^n$ and $\begin{vmatrix} 3 & 1 + f(1) & 1 + f(2) \\ 1 + f(1) & 1 + f(2) & 1 + f(3) \\ 1 + f(2) & 1 + f(3) & 1 + f(4) \end{vmatrix} = K(1 - \alpha)^2 (1 - \beta)^2 (\alpha - \beta)^2$,then $K = \dots$

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